Mathematics
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OpenStudy (anonymous):
can someone please help me with four Rational Expressions?
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OpenStudy (anonymous):
here are the questions
OpenStudy (hba):
\[\frac{ x }{ 3x }+\frac{ 2x-3 }{ 3x }\]
Start by taking lcm.
OpenStudy (hba):
@Spartan_Of_Ares
Start solving,don't worry i am here with you :)
OpenStudy (anonymous):
whats Icm
OpenStudy (hba):
For example
1/2+1/2
1 + 1
=-------
2
= 2/2
=1
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OpenStudy (hba):
@Spartan_Of_Ares
Understood ,now take lcm :)
OpenStudy (anonymous):
so then \[x+2x-3/3x\]
OpenStudy (hba):
Yeah and add the things which can be added .
OpenStudy (anonymous):
what can be added?
OpenStudy (hba):
x+2x can be added.
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OpenStudy (anonymous):
so i would get 3x or 2x^2
OpenStudy (hba):
You will get 3x
so do that and show me what you got.
OpenStudy (anonymous):
so i would get 3x-3/3x?
OpenStudy (anonymous):
and then cancel 3x and 3x?
OpenStudy (hba):
Yeah right for the first comment
No lol for the 2nd comment
Take common from the upper thing first
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OpenStudy (anonymous):
i dont know what you mean
OpenStudy (hba):
\[\frac{ 3x-3 }{ 3 },now \ you \ can \ take \ common \]
OpenStudy (anonymous):
so 3 from the top and three from bottom?
OpenStudy (hba):
But my mistake it was
\[3x-3/3x\]
and yeah you are right :)
So what do you get ?
OpenStudy (anonymous):
3x/x
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OpenStudy (hba):
@Spartan_Of_Ares
Gone mad ?
I told you to take common first.
Show me the steps.
OpenStudy (anonymous):
x+2x-3/3x --> 3x-3/3x --> 3?
OpenStudy (hba):
You are correct till 3x-3/3x
now you have to take common from the numerator
3(x-1)/3x
and then cancel
OpenStudy (anonymous):
so i get x-1/3?
OpenStudy (anonymous):
i mean x-1/x
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OpenStudy (hba):
What ?
OpenStudy (hba):
Yeah the second one is right
OpenStudy (hba):
x-1/x is correct
Bye :)
OpenStudy (anonymous):
bye and thanks
OpenStudy (hba):
Welcome :)